How do you solve #0.23x^2+6.5x+4.3<0# using a sign chart?

1 Answer
Nov 15, 2017

Answer:

Solution : # -27.58 < x < -0.68 or x| (-27.58,-0.68)#

Explanation:

#0.23x^2+6.5x+4.3<0#

Comparing with standard quadratic equation #ax^2+bx+c=0#

# a= 0.23 ,b=6.5 ,c=4.3# Discriminant # D= b^2-4ac# or

#D ~~ 38.29# If discriminant positive, we get two real solutions,

Quadratic formula: #x= (-b+-sqrtD)/(2a) #or

#x= (-6.5+-sqrt38.29)/(2*0.23) :. x ~~ -27.58 , x ~~ -0.68#

#0.23x^2+6.5x+4.3<0 #or

# f(x)=0.23 (x +27.58)(x+0.68) <0 # .

Critical points are # x ~~ -27.58 , x ~~ -0.68#

Sign chart: When #x< -27.58# sign of #f(x) # is # (-) * (-) = (+) ; > 0#

When # -27.58 < x < -0.68 # sign of #f(x) # is # (+) * (-) = (-) ; < 0#

When #x > -0.68# sign of #f(x) # is # (+) * (+) = (+) ; > 0#

Solution : # -27.58 < x < -0.68 or x| (-27.58,-0.68)#

graph{0.23x^2+6.5x+4.3 [-160, 160, -80, 80]}

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